Consider a body of mass #m# on an incline with an inclination angle of #\theta#. To this a force is applied without disturbing the equilibrium. The forces acting on the object are
#vec W # : Weight of the object acting vertically downward;
#vec F_{app} # : Force applied parallel to the inclined plane,
#vecF_f# : Frictional force acting parallel to the plane of the incline,
#vec N # : Normal force acting perpendicular to the plane of the incline,
Consider a coordinate system with its X-axis parallel to the plane downward and Y-axis perpendicular to the plane upward.
Resolve the weight into components that are parallel and perpendicular to the inclined plane
#vec W_{||} = -mg\sin\theta; \qquad vec W_{\_|_} = -mg\cos\theta#
Equilibrium: #\qquad vec W + vec F_{app} + vec N + vec F_f = vec 0#
(A) Perpendicular Component: #\qquad vec N + vec W_{\_|_} = vec 0#
#N - W_{\_|_} = 0; \qquad N = mg\cos\theta;#
(B) Parallel Component: #\qquad vec W_{||} + F_{app} + vec F_f = vec 0#
#W_{||} + F_{app} - F_f = 0; \qquad F_f = F_{app} + W_{||} # ...... (2)
Static Equilibrium: In this case the frictional force is due to static friction.
#F_f = \mu_sN = \mu_s.mg\cos\theta; \qquad F_{app} = F_{app}^s = 15# #N;#
Expanding (2), #\mu_s.mg\cos\theta = F_{app}^s + mg\sin\theta#
#\mu_s = (F_{app}^s/(mg\cos\theta) + \tan\theta)# ...... (3)
Dynamic Equilibrium: In this case the frictional force is due to kinetic friction.
#F_f = \mu_kN = \mu_k.mg\cos\theta; \qquad F_{app} = F_{app}^k = 5# #N;#
Expanding (2), #\mu_k.mg\cos\theta = F_{app}^k + mg\sin\theta#
#\mu_k = (F_{app}^k/(mg\cos\theta) + \tan\theta)# ...... (4)
Calculate:
#\sin\theta = \cos\theta = 1/\sqrt{2}; \qquad \tan\theta = 1; \qquad mg\cos\theta = 62.37# #N#
#mu_s = 1.24; \qquad \mu_k = 1.08#
